A072726 Numerator of the rationals >= 1 whose continued fractions consist of only even terms, in ascending order by the sum of the continued fraction terms and descending by lowest order continued fraction terms to highest.

1, 2, 4, 5, 6, 9, 9, 12, 8, 13, 17, 22, 13, 20, 22, 29, 10, 17, 25, 32, 25, 38, 40, 53, 17, 28, 38, 49, 32, 49, 53, 70, 12, 21, 33, 42, 37, 56, 58, 77, 33, 54, 72, 93, 58, 89, 97, 128, 21, 36, 54, 69, 56, 85, 89, 118, 42, 69, 93, 120, 77, 118, 128, 169

Offset: 0

Author

Paul D. Hanna, Jul 09 2002

Keywords

• easy
• frac
• nice
• nonn

Examples

			n: a(n)/A072727 has continued fraction:
0: 1/0 = [infinity]
1: 2/1 = [2]
2: 4/1 = [4]
3: 5/2 = [2;2]
4: 6/1 = [6]
5: 9/2 = [4;2]
6: 9/4 = [2;4]
7: 12/5 = [2;2,2]
8: 8/1 = [8]
9: 13/2 = [6;2]
10: 17/4 = [4;4]
11: 22/5 = [4;2,2]
12: 13/6 = [2;6]
13: 20/9 = [2;4,2]
14: 22/9 = [2;2,4]
15: 29/12= [2;2,2,2]
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..1023

Crossrefs

Cf. A072727, A071585, A071766, A072728, A072729, A000129.

Programs

Mathematica

a[0] = 1; a[n_] := a[n] = Which[IntegerQ[k = Log[2, n]], 2 (k + 1), IntegerQ[k = Log[2, n - 1]], 4 k + 1, IntegerQ[k = Log[2, n + 1]], Fibonacci[k + 1, 2], True, Clear[k]; Hold[2*(k - j)*a[2^j + m] + a[m]] /. ToRules[Reduce[2^k > 2^j > m >= 0 && n == 2^k + 2^j + m, {k, j, m}, Integers]] // ReleaseHold];
Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 63}] (* Jean-François Alcover, Jul 13 2016 *)

Formula

a(2^k + 2^j + m) = 2(k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >=0. a(0) = 1, a(2^k) = 2(k+1), a(2^k + 1) = 4*k + 1 (k>0), a(2^k - 1) = the (k+1)-th Pell number.